Differentiate:
y= ln(x-1/x+1)
answer is:
2/x^2 - 1
tried the quotient rule but not sure how to get the right answer
2007-04-08 07:29:04 · 7 answers · asked by xsouthern_belle 1 in Science & Mathematics ➔ Mathematics
please include step by step
2007-04-08 07:32:52 · update #1
y = ln ( (x-1)/(x+1) )
= ln(x-1) - ln(x+1)
y' = 1/(x-1) - 1/(x+1)
= [ (x+1) - (x-1) ] / [(x-1)(x+1)]
= 2/(x^2-1)
2007-04-08 07:36:17 · answer #1 · answered by hustolemyname 6 · 0⤊ 0⤋
Calculus Problem?
Differentiate:
y= ln(x-1/x+1)=log(x-1-log(x+1
its derivative =1/(x-1) -1/(x+1) =2/x^2_-1) answer
2007-04-08 14:42:50 · answer #2 · answered by Anonymous · 0⤊ 0⤋
y'=(x+1/x-1)(1(x+1)-1(x-1))/(x+1)^2
= (x+1)2/(x-1)(x+1)^2
=2/(x+1)(x-1)=
2/(x^2-1)
2007-04-08 14:37:18 · answer #3 · answered by murnau 3 · 0⤊ 0⤋
Easier if you separate it first.
y = ln(x-1) - ln(x+1)
y' = 1/(x-1) - 1/(x+1)
Then when you subtract the fractions you get the answer you have.
2007-04-08 14:36:16 · answer #4 · answered by David K 3 · 0⤊ 0⤋
y = ln ( (x-1)/(x+1) )
y= ln(x-1) - ln(x+1)
dy/dx= 1/(x-1) - 1/(x+1)
= [ (x+1) - (x-1) ] / [(x-1)(x+1)]
= x+1-x+1 / [(x-1)(x+1)]
=2/ ((x^2-1))
2007-04-08 15:15:21 · answer #5 · answered by Khalidxp 3 · 0⤊ 0⤋
d(lnu) = u'/u
so 1/u = (x+1) /(x-1)
u' = (x+1)-(x-1) /(x+1)^2 = 2/(x+1)^2
d(lnu) =2 (x+1) /(x-1)/
u' = 2(x+1)-(x-1) /(x+1)^2
you simplify by (x+1)
ypu find the answer 2/(x^2-1)
2007-04-08 14:37:27 · answer #6 · answered by maussy 7 · 0⤊ 0⤋
http://www.amazon.com/s/ref=nb_ss_gw/102-6986154-3624919?url=search-alias%3Daps&field-keywords=schaum%27s+calculus
2007-04-08 14:33:30 · answer #7 · answered by bobweb 7 · 0⤊ 1⤋